页面加载菜单栏上的JavaScript活动类

Question

我使用 Laravel 构建了一个管理面板。我有多个菜单的侧边栏。我面临的问题是，每当我单击其中一个菜单时，我都想激活该类，但它无法保持该活动状态。

这是我侧边栏的代码

<div id="sidebar">   
    <ul>
        <li class="submenu"> <a href="#"><i class="icon icon-list"></i> <span>Members</span></a>
            <ul>
              <li class="submenu-2"><a href="{{asset('member/create')}}">Create Member</a></li>          
              <li class="submenu-2"><a href="{{asset('member')}}">Member List</a></li>               
            </ul>
        </li>
        <li class="submenu"> <a href="#"><i class="icon icon-list"></i> <span>Payroll</span></a>
            <ul>
                    <li class="submenu-2"><a href="{{asset('payroll/create')}}">Create Monthly Payroll</a></li>               
                    <li class="submenu-2"><a href="{{asset('managerPayroll')}}">Unapproved Monthly Payroll (Management)</a></li>                           
                    <li class="submenu-2"><a href="{{asset('nonManagerPayroll')}}">Unapproved Monthly Payroll (Employee)</a></li>
                    <li class="submenu-2"><a href="{{asset('printEmployeeAccount')}}">Print Employee Account</a></li>                     
            </ul>
        </li>
        <li class="submenu"> <a href="#"><i class="icon icon-list"></i> <span>Report</span></a>
            <ul>
                    <li class="submenu-2"><a href="{{asset('serachManagerReport')}}">Search Manager Report</a></li> 
                    <li class="submenu-2"><a href="{{asset('serachNonManagerReport')}}">Search Employee Report</a></li>
                    <li class="submenu-2"><a href="{{asset('salaryReport')}}">Salary Sheet </a></li> 
            </ul>
        </li>
        <li class="submenu-2"> <a href="{{asset('changePassword')}}"><i class="icon icon-list"></i> <span>Password Update</span></a>
        </li>
    </ul>
</div> 

以及我用来使网址处于活动状态的javascript。

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>         
<script>
    (function() {
    var nav = document.getElementById('sidebar'),
        anchor = nav.getElementsByTagName('a'),
        current = window.location.pathname.split('/')[1];
        for (var i = 0; i < anchor.length; i++) {
        if(anchor[i].href == current) {
            anchor[i].className = "active";
        }
    }
})();
</script>

任何人都可以建议这里的错误是什么吗？

Answer 1

您可以通过简单的方式执行此操作：按以下方式在每个菜单中添加@yield：

<li class="@yield('menu_create_member') submenu-2"><a href="{{asset('member/create')}}">Create Member</a></li>

对于主菜单和子菜单，请尝试以下操作：

<li class="@yield('menu_member') submenu"> <a href="#"><i class="icon icon-list"></i> <span>Members</span></a>
  <ul>
   <li class="@yield('menu_create_member') submenu-2"><a href="{{asset('member/create')}}">Create Member</a></li>                        
   </ul>
 </li>

然后在扩展主模板后(@extends('layouts.master'))加载时将其添加到每个页面中

@section('menu_member','active')  // For main menu

@section('menu_create_member','active') // For sub menu

然后自动激活当前菜单

Answer 2

像这样$。加载 jquery。或者您也可以使用$(document).ready(function(){});

$(function() {
    var nav = document.getElementById('sidebar'),
        anchor = nav.getElementsByTagName('a'),
        current = window.location.pathname.split('/')[1];
        for (var i = 0; i < anchor.length; i++) {
        if(anchor[i].href == current) {
            anchor[i].className = "active";
        }
    }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="sidebar">   
    <ul>
        <li class="submenu"> <a href="#"><i class="icon icon-list"></i> <span>Members</span></a>
            <ul>
              <li class="submenu-2"><a href="{{asset('member/create')}}">Create Member</a></li>          
              <li class="submenu-2"><a href="{{asset('member')}}">Member List</a></li>               
            </ul>
        </li>
        <li class="submenu"> <a href="#"><i class="icon icon-list"></i> <span>Payroll</span></a>
            <ul>
                    <li class="submenu-2"><a href="{{asset('payroll/create')}}">Create Monthly Payroll</a></li>               
                    <li class="submenu-2"><a href="{{asset('managerPayroll')}}">Unapproved Monthly Payroll (Management)</a></li>                           
                    <li class="submenu-2"><a href="{{asset('nonManagerPayroll')}}">Unapproved Monthly Payroll (Employee)</a></li>
                    <li class="submenu-2"><a href="{{asset('printEmployeeAccount')}}">Print Employee Account</a></li>                     
            </ul>
        </li>
        <li class="submenu"> <a href="#"><i class="icon icon-list"></i> <span>Report</span></a>
            <ul>
                    <li class="submenu-2"><a href="{{asset('serachManagerReport')}}">Search Manager Report</a></li> 
                    <li class="submenu-2"><a href="{{asset('serachNonManagerReport')}}">Search Employee Report</a></li>
                    <li class="submenu-2"><a href="{{asset('salaryReport')}}">Salary Sheet </a></li> 
            </ul>
        </li>
        <li class="submenu-2"> <a href="{{asset('changePassword')}}"><i class="icon icon-list"></i> <span>Password Update</span></a>
        </li>
    </ul>
</div>

Answer 3

通过将代码包装在括号中，您不会在里面调用函数。

(function() {
    var nav = document.getElementById('sidebar'),
        anchor = nav.getElementsByTagName('a'),
        current = window.location.pathname.split('/')[1];
    for (var i = 0; i < anchor.length; i++) {
        if(anchor[i].href == current) {
            anchor[i].className = "active";
            break;
        }
    }
}) (); // Add 2 parentheses to call the anonymous function

附言。我还在 for 循环中添加了一个中断，因为我想你只需要让每页激活一个链接